Michael Feathers wrote:
I wrote this function the other day, and I was wondering if I'm missing
something.. whether there is already a function or idiom around to do this.
unlist3 :: (a -> a -> a -> b) -> [a] -> b
unlist3 f (x:y:z:xs) = f x y z
I was also wondering whether the function can be generalized to N or
whether this is just one of those edges in the type system that you
can't abstract over.
In Scheme and Lisp, the "apply" function does this, so the following
Scheme definition would do the trick:
(define unlist apply)
(Although this doesn't ignore the xs as in the example; that would
require some code to extract a sublist of the desired length.)
In Haskell, bring on the type system shenanigans! ;)
Anton
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